XI Mathematics IIT JEE Advanced Study Package 2014 159

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BRILLIANT PUBLIC SCHOOL,

SITAMARHI

(Affiliated up to +2 level to C.B.S.E., New Delhi)

Class-XI

IIT-JEE Advanced Mathematics

Study Package

Session: 2014-15

Office: Rajopatti, Dumra Road, Sitamarhi (Bihar), Pin-843301

Ph.06226-252314 , Mobile:9431636758, 9931610902

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STUDY PACKAGE

Target: IIT-JEE (Advanced)

SUBJECT: MATHEMATICS-XI

Chapter

Pages Exercises

1 Trigonometric Ratio and Identity

19

5 2 Trigonometric Equations

14

3 3 Properties of Triangle

24

5 4 Functions

40

5 5 Complex Numbers

37

5 6 Quadratic Equations

23

6 7 Permutations and Combinations

19

5 8 Binomial Theorem

24

8 9 Probability

36

5 10 Progressions

25

5 11 Straight Lines

21

5 12 Circles

23

5 13 Parabola, Ellipse and Hyperbola

68

15 14 Highlights on Conic Sections

25 15 Vector Algebra and 3-D Geometry

62

8 16 Limits

18

5

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STUDY PACKAGE

Target: IIT-JEE (Advanced)

SUBJECT: MATHEMATICS

TOPIC: 1 XI M 1. Trigonometric

Ratio and Identity

Index:

1. Key Concepts

2. Exercise I to V

3. Answer Key

4. Assertion and Reasons

5. 34 Yrs. Que. from IIT-JEE

6. 10 Yrs. Que. from AIEEE

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Trigonometric Ratios

& Identities

1 .

Basic Trigonometric Identities:

(a) sin²θ + cos²θ = 1; −1 ≤ sin θ ≤ 1; −1 ≤ cos θ ≤ 1 ∀ θ ∈ R (b) sec²θ − tan²θ = 1 ; secθ ≥ 1 ∀ θ ∈ R –

(

)

      Ι ∈ π + ,n 2 1 n 2

(c) cosec²θ − cot²θ = 1 ; cosecθ ≥ 1 ∀ θ ∈ R –

{

nπ n, ∈Ι

}

Solved Example # 1

Prove that

(i) cos4_{A – sin}4_{A + 1 = 2 cos}2_A

(ii) 1 A sec A tan 1 A sec A tan + − − + = A cos A sin 1+ Solution

(i) cos4_{A – sin}4_{A + 1}

= (cos2_{A – sin}2_{A) (cos}2_{A + sin}2_{A) + 1}

= cos2_{A – sin}2_{A + 1} [∴ cos2_{A + sin}2_{A = 1]}

= 2 cos2_A (ii) 1 A sec A tan 1 A sec A tan + − − + = 1 A sec A tan ) A tan A (sec A sec A tan 2 2 + − − − + = 1 A sec A tan ) A tan A sec 1 )( A sec A (tan + − + − + = tan A + sec A = A cos A sin 1+ Solved Example # 2

If sin x + sin2_{x = 1, then find the value of}

cos12_{x + 3 cos}10_{x + 3 cos}8_{x + cos}6_{x – 1}

Solution

cos12_{x + 3 cos}10_{x + 3 cos}8_{x + cos}6_{x – 1}

= (cos4_{x + cos}2_x)3_{– 1}

= (sin2_{x + sinx)}3_{– 1} _{[∵ cos}2_{x = sin x]}

= 1 – 1 = 0 Solved Example # 3 If tan θ = m – m 4 1

, then show that sec θ – tan θ = – 2m or m 2

1 Solution

Depending on quadrant in which θ falls, sec θ can be ± m 4 1 m 4 2+ So, if sec θ = 4m2+1 = m + 1 2

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⇒ _{sec θ – tan θ =} m 2 1 and if sec θ = –       + m 4 1 m ⇒ _{sec θ – tan θ = – 2m}

Self Practice Problem

1. Prove the followings :

(i) cos6_{A + sin}6_{A + 3 sin}2_{A cos}2_{A = 1}

(ii) sec2_{A + cosec}2_{A = (tan A + cot A)}2

(iii) sec2_{A cosec}2_{A = tan}2_{A + cot}2_{A + 2}

(iv) (tan α + cosec β)2 – (cot β – sec α)2 = 2 tan α cot β (cosec α + sec β)

(v) _ 2_α₋ 2_α+_cos_ec2_α₋_sin2_α_ 1 cos sec 1 cos2α sin2α = α α + α α − 2 2 2 2 cos sin 2 cos sin 1 2. If sin θ = ₂ ₂ 2 n 2 mn 2 m mn 2 m + + +

, then prove that tan θ = 2 ₂

n 2 mn 2 mn 2 m + +

2 .

C i r c u l ar

D e f i n i t i o n

O f

T r i g o no me t r i c

F u n c t i o n s :

sin θ = OP PM cos θ = OP OM

tan θ =_cossinθ_θ, cos θ ≠ 0 cot θ =cos_sin_θθ, sin θ ≠ 0

sec θ = _θ cos 1 , cos θ ≠ 0 cosec θ = _θ sin 1 , sin θ ≠ 0

3 .

T r i g o no me t r i c

F u n c t io n s

O f

A l l i ed

A n g l e s :

If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES.

(a) sin (− θ) = − sinθ ; cos (− θ) = cosθ

(b) sin (90° − θ) = cosθ ; cos (90°− θ) = sinθ (c) sin (90° + θ) = cosθ ; cos (90° + θ) = − sinθ (d) sin (180° − θ) = sinθ ; cos (180° − θ) = − cosθ (e) sin (180° + θ) = − sinθ ; cos (180° + θ) = − cosθ (f) sin (270° − θ) = − cosθ ; cos (270° − θ) = − sinθ (g) sin (270° + θ) = − cosθ ; cos (270° + θ) = sinθ

(h) tan (90° − θ) = cotθ ; cot (90° − θ) = tanθ

Solved Example # 4 Prove that

(i) cot A + tan (180º + A) + tan (90º + A) + tan (360º – A) = 0 (ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1 = 0 Solution

(i) cot A + tan (180º + A) + tan (90º + A) + tan (360º – A) = cot A + tan A – cot A – tan A = 0

(ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1 = – cosec2_{A + cot}2_{A + 1 = 0}

3. Prove that

(i) sin 420º cos 390º + cos (–300º) sin (–330º) = 1

(ii) tan 225º cot 405º + tan 765º cot 675º = 0

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4

4 4 ....

Graphs of Trigo nometric functio ns:

(a) y = sin x x ∈ R; y ∈ [–1, 1] (b) y = cos x x ∈ R; y ∈ [ – 1, 1] (c) y = tan x x ∈ R – (2n + 1) π/2, n ∈ Ι ; y ∈ R (d) y = cot x x ∈ R – nπ , n ∈ Ι; y ∈ R (e) y = cosec x x ∈ R – nπ , n ∈ Ι ; y ∈ (− − − − ∞, −−−− 1] ∪∪∪∪ [1, ∞∞∞)∞ (f) y = sec x x ∈ R – (2n + 1) π/2, n ∈ Ι ; y ∈∈∈∈ (− ∞, − 1] ∪ [1, ∞)

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Find number of solutions of the equation cos x = |x| Solution

Clearly graph of cos x & |x| intersect at two points. Hence no. of solutions is 2 Solved Example # 6

Find range of y = sin2x + 2 sin x + 3 ∀ x ∈ R

Solution

We know – 1 ≤ sin x ≤ 1

⇒ 0 ≤ sin x +1 ≤ 2

⇒ _{2 ≤ (sin x +1)}2 + 2 ≤ 6

Hence range is y ∈ [2, 6] Self Practice Problem

4. Show that the equation sec2θ =

2 ) y x ( xy 4

+ is only possible when x = y ≠ 0

5. Find range of the followings.

(i) y = 2 sin2x + 5 sin x +1∀ x ∈ R _Answer _{[–2, 8]}

(ii) y = cos2_{x – cos x + 1} ∀ x ∈ R _Answer _

     3 , 4 3

6. Find range of y = sin x, x ∈ 

     π π 2 3 2 Answer _       − 2 3 , 1

5

5 5 ....

Trigonometric Functions of Sum or Difference of Two Angles:

(a) sin (A ± B) = sinA cosB ± cosA sinB

(b) cos (A ± B) = cosA cosB ∓ sinA sinB

(c) sin²A − sin²B = cos²B − cos²A = sin (A+B). sin (A− B)

(d) cos²A − sin²B = cos²B − sin²A = cos (A+B). cos (A− B)

(e) tan(A ± B) = ₁tan_tanA_Atan_tanB_B ∓

±

(f) cot (A ± B) =cot_cotA_Bcot_±_cotB_A1 ∓

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(g) tan (A + B + C) =₁tan₋_tanA_A+tan_tanB_B+₋tan_tanC_B−_tantan_CA₋tan_tanB_Ctan_tanC_A .

Solved Example # 7 Prove that

(i) sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B)

(ii) tan       θ + π 4 tan      θ + π 4 3 = –1 Solution

(i) Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = sin (45º + A + 45º – B)

= sin (90º + A – B) = cos (A – B)

(ii) tan _π₄+θ_ × tan _3₄π+θ_

= ₋+ _θθ tan 1 tan 1 × −₊+ _θθ tan 1 tan 1 = – 1 Self Practice Problem

7. If sin α = 5 3 , cos β = 13 5

, then find sin (α + β) Answer –

65 33 ,

65 63

8. Find the value of sin 105º Answer

2 2

1 3+

9. Prove that 1 + tan A tan

2 A = tan A cot 2 A – 1 = sec A

6

6 6 ....

F a c t o r i s at i o n o f t h e S u m o r D i f f e r e nc e o f T w o S i ne s o r

C o s i n e s :

(a) sinC + sinD = 2 sin

2 D C+ cos 2 D C−

(b) sinC − sinD = 2 cos

2 D C+ sin 2 D C−

(c) cosC + cosD = 2 cos

2 D C+ cos 2 D C−

(d) cosC−cosD = −2 sin

2 D C+ sin 2 D C− Solved Example # 8

Prove that sin 5A + sin 3A = 2sin 4A cos A Solution

L.H.S. sin 5A + sin 3A = 2sin 4A cos A = R.H.S.

[∵ sin C + sin D = 2 sin 2 D C+ cos 2 D C− ] Solved Example # 9

Find the value of 2 sin 3θ cos θ – sin 4θ – sin 2θ Solution

2 sin 3θ cos θ – sin 4θ – sin 2θ = 2 sin 3θ cos θ – [2 sin 3θ cos θ ] = 0 Self Practice Problem

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10. Proved that

(i) cos 8x – cos 5x = – 2 sin

2 x 13 sin 2 x 3

(ii) _cossin_AA₋+_cossin2₂A_A = cot 2 A (iii) A 7 cos A 5 cos A 3 cos A cos A 7 sin A 5 sin A 3 sin A sin + + + + + + = tan 4A (iv) A 7 sin A 5 sin 2 A 3 sin A 5 sin A 3 sin 2 A sin + + + + = A 5 sin A 3 sin (v) A 13 cos A 9 cos A 5 cos A cos A 13 sin A 9 sin A 5 sin A sin + − − − + − = cot 4A

7

7 7....

Tr ansfo r mat io n o f P r od uc ts into S um o r D if fer en ce of S in es &

C o s i n e s

C o s i n e s ::::

(a) 2 sinA cosB = sin(A+B) + sin(A−B) (b) 2 cosA sinB = sin(A+B) − sin(A−B)

(c) 2 cosA cosB = cos(A+B) + cos(A−B) (d) 2 sinA sinB = cos(A−B) − cos(A+B)

Solved Example # 10 Prove that (i) θ_θ θ_θ−₋ θ_θ θ_θ 4 sin 3 sin cos 2 cos 3 cos 6 sin cos 8 sin = tan 2θ (ii) _θθ₋+ _θθ 3 tan 5 tan 3 tan 5 tan = 4 cos 2θ cos 4θ Solution (i) θ_θ θ_θ−₋ θ_θ _θθ 4 sin 3 sin 2 cos 2 cos 2 3 cos 6 sin 2 cos 8 sin 2 = θ_θ+₊ _θθ₋− _θθ₊− _θθ 7 cos cos cos 3 cos 3 sin 9 sin 7 sin 9 sin = θ_θ θ_θ 2 cos 5 cos 2 5 cos 2 sin 2 = tan 2θ (ii) _θθ₋+ _θθ 3 tan 5 tan 3 tan 5 tan = _θθ _θθ₋+ _θθ _θθ 5 cos 3 sin 3 cos 5 sin 5 cos 3 sin 3 cos 5 sin = _θθ 2 sin 8 sin = 4 cos2θ cos 4θ Self Practice Problem

11. Prove that sin

2 θ sin 2 7θ + sin 2 3θ sin 2 11θ = sin 2θ sin 5θ

12. Prove that cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B) = 0

13. Prove that 2 cos

13 π cos 13 9π + cos 13 3π + cos 13 5π = 0

8

8 8....

Multiple and Sub-multiple Angles :

(a) sin 2A = 2 sinA cosA ; sinθ = 2 sinθ

2 cos θ 2

(b) cos 2A = cos²A − sin²A = 2cos²A−1 = 1 − 2 sin²A; 2 cos²

2 θ = 1 + cosθ, 2 sin² 2 θ = 1 − cosθ. (c) tan 2A = A tan 1 A tan 2 2 − ; tanθ = 2₂ 2 tan 1 tan 2 θ θ − (d) sin 2A = A tan 1 A tan 2 2 + , cos 2A =1 tan A A tan 1 2 2 + − 7

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(e) sin 3A = 3 sinA− 4 sin3_A _(f) _{cos 3A = 4 cos}3A − 3 cosA (g) tan 3A = A tan 3 1 A tan A tan 3 2 3 − − Solved Example # 11 Prove that (i) A 2 cos 1 A 2 sin + = tan A

(ii) tan A + cot A = 2 cosec 2 A

(iii) ) B A cos( B cos A cos 1 ) B A cos( B cos A cos 1 + − − + + − + − = tan 2 A cot 2 B Solution (i) L.H.S. A 2 cos 1 A 2 sin + = 2cos A A cos A sin 2 2 = tan A

(ii) L.H.S. tan A + cot A =

A tan A tan 1+ 2 = 2        + A tan 2 A tan 1 2 = A 2 sin 2 = 2 cosec 2 A (iii) L.H.S. ) B A cos( B cos A cos 1 ) B A cos( B cos A cos 1 + − − + + − + − =       + −       + + B 2 A cos 2 A cos 2 2 A cos 2 B 2 A sin 2 A sin 2 2 A sin 2 2 2 = tan 2 A                   + −       + + B 2 A cos 2 A cos B 2 A sin 2 A sin = tan 2 A                   +       + 2 B sin 2 B A sin 2 2 B cos 2 B A sin 2 = tan 2 A cot 2 B

14. Prove that ₊ θ_θ+₊ θ _θ 2 cos cos 1 2 sin sin = tan θ

15. Prove that sin 20º sin 40º sin 60º sin 80º = ₁₆3

16. Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A

17. Prove that tan _45º+A₂_ = sec A + tan A

9

9 9 ....

Important Trigonometric Ratios:

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(b) sin 15° or sin 12 π = 2 2 1 3− = cos 75° or cos 12 5π ; cos 15° or cos 12 π = 2 2 1 3+ = sin 75° or sin 12 5π ; tan15° = 1 3 1 3 + − =2− 3 = cot 75° ; tan75° = 1 3 1 3 − + =2+ 3 = cot 15° (c) sin 10 π or sin 18° = 4 1 5−

& cos 36° or cos 5 π = 4 1 5+

1 0 .

1 0 . C o n d i t i o na l

C o n d i t i o na l

I d e n t i t i e s

I d e n t i t i e s ::::

If A + B + C = π then :

(i) sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii) sinA + sinB + sinC = 4 cos

2 A cos 2 B cos 2 C

(iii) cos 2A + cos 2B + cos 2C = − 1 − 4 cos A cos B cos C

(iv) cos A + cos B + cos C = 1 + 4 sin

2 A sin 2 B sin 2 C

(v) tanA + tanB + tanC = tanA tanB tanC

(vi) tan 2 A tan 2 B + tan 2 B tan 2 C + tan 2 C tan 2 A = 1 (vii) cot 2 A + cot 2 B + cot 2 C = cot 2 A . cot 2 B . cot 2 C (viii) cot A cot B + cot B cot C + cot C cot A = 1

(ix) A + B + C =π

2 then tan A tan B + tan B tan C + tan C tan A = 1 Solved Example # 12

If A + B + C = 180°, Prove that, sin2_{A + sin}2_{B + sin}2_{C = 2 + 2cosA cosB cosC.}

Solution.

Let S = sin2_{A + sin}2_{B + sin}2_C

so that 2S = 2sin2_{A + 1 – cos2B +1 – cos2C}

= 2 sin2_{A + 2 – 2cos(B + C) cos(B – C)}

= 2 – 2 cos2_{A + 2 – 2cos(B + C) cos(B – C)}

∴ S = 2 + cosA [cos(B – C) + cos(B+ C)]

since cosA = – cos(B+C)

∴ S = 2 + 2 cos A cos B cos C

If x + y + z = xyz, Prove that ₂

x 1 x 2 − + ₁ _y2 y 2 − + ₁ _z2 z 2 − = ₁ _x2 x 2 − . 2 y 1 y 2 − . ₁ _z2 z 2 − . Solution.

Put x = tanA, y = tanB and z = tanC,

so that we have

tanA + tanB + tanC = tanA tanB tanC ⇒ A + B + C = nπ, where n ∈ Ι

Hence L.H.S.

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∴ ₁ _x2 x 2 − + ₁ _y2 y 2 − + ₁ _z2 z 2 − = 1 tan A A tan 2 2 − + 1 tan B B tan 2 2 − + 1 tan C C tan 2 2 − .

= tan2A + tan2B + tan2C [[[[∵ A + B + C = nπ ]

= tan2A tan2B tan2C = ₁ _x2 x 2 − . 1 y2 y 2 − . ₁ _z2 z 2 − Self Practice Problem

18. If A + B + C = 180°, prove that

(i) sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = 4sin

2 C B− sin 2 A C− sin 2 B A− (ii) C sin B sin A sin C 2 sin B 2 sin A 2 sin + + + + = 8 sin 2 A sin 2 B sin 2 C . 19. If A + B + C = 2S, prove that

(i) sin(S – A) sin(S – B) + sinS sin (S – C) = sinA sinB.

(ii) sin(S – A) + sin (S – B) + sin(S – C) – sin S = 4sin 2 A sin 2 B sin 2 C .

1 1

1 1 .... Range

Range

o

o f

f

Trigonometric

Expression

Expression::::

E = a sinθ + b cosθ

E = 2 2

b

a + sin (θ + α), where tan α =_ab

= _a2₊_b2 _{cos (θ − β), where tan β =}

b a

Hence for any real value of θ, ₋ _a2₊_b2 _≤ _E _≤ _a2₊_b2

Find maximum and minimum values of following :

(i) 3sinx + 4cosx

(ii) 1 + 2sinx + 3cos2_x

Solution.

(i) We know

– 32+42 ≤ 3sinx + 4cosx ≤ 32+42 – 5 ≤ 3sinx + 4cosx ≤ 5

(ii) 1+ 2sinx + 3cos2_x

= – 3sin2_{x + 2sinx + 4} = – 3 _sin2x−2sin₃ x_ + 4 = – 3 2 3 1 x sin       − ₊ 3 13 Now 0 ≤ 2 3 1 x sin       − _≤ 9 16 ⇒ _– 3 16 ≤ – 3 2 3 1 x sin       − _{≤ 0}

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– 1 ≤ – 3 2 3 1 x sin       − ₊ 3 13 ≤ 3 13

20. Find maximum and minimum values of following

(i) 3 + (sinx – 2)2 _Answer _{max = 12, min = 4.}

(ii) 10cos2_{x – 6sinx cosx + 2sin}2_x _Answer _{max = 11, min = 1.}

(iii) cosθ + 3 2sin _θ+π₄_ + 6 Answer max = 11, min = 1

1

1 1 2

2

2 2 .... Sine

Sine

Sine a

a

a nd Cosine Series

nd Cosine Series

nd Cosine Series::::

nd Cosine Series

sinα + sin(α + β) + sin(α + 2β ) +... + sin

(

α+n−1β

)

= 2 2 n sin sin β β sin_α+ n₂−1β_

cosα + cos(α + β) + cos(α + 2β ) +... + cos

(

α+n−1β

)

= 2 2 n sin sin β β cos _     β − + α 2 1 n Solved Example # 15

Find the summation of the following

(i) cos 7 2π + cos 7 4π + cos 7 6π (ii) cos 7 π + cos 7 2π + cos 7 3π + cos 7 4π + cos 7 5π + cos 7 6π (iii) cos 11 π + cos 11 3π + cos 11 5π + cos 11 7π + cos 11 9π Solution. (i) cos 7 2π + cos 7 4π + cos 7 6π = 7 sin 7 3 sin 2 7 6 7 2 cos π π       π₊ π = 7 sin 7 3 sin 7 4 cos π π π = 7 sin 7 3 sin 7 3 cos π π π − = – 7 sin 2 7 6 sin π π = – 2 1 (ii) cos 7 π + cos 7 2π + cos 7 3π + cos 7 4π + cos 7 5π + cos 7 6π 11

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= 14 sin 14 6 sin 2 7 6 7 cos π π            π₊ π = 14 sin 14 6 sin 2 cos π π π = 0 (iii) cos 11 π + cos 11 3π + cos 11 5π + cos 11 7π + cos 11 9π = 11 sin 11 5 sin 22 10 cos π π π = 11 sin 2 11 10 sin π π = 2 1

Find sum of the following series :

21. cos 1 n 2 + π + cos 1 n 2 3 + π + cos 1 n 2 5 + π + ... + to n terms. Answer 2 1

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SHORT REVISION

Trigonometric Ratios & Identities

1 .

B

ASIC

T

RIGONOMETRIC

I

DENTITIES

:

(a)sin

2

θ + cos

2

θ = 1 ;

−1 ≤ sin θ ≤ 1 ;

−1 ≤ cos θ ≤ 1 ∀ θ ∈ R

(b)sec

2

θ − tan

2

θ = 1 ;

sec

θ ≥ 1 ∀ θ ∈ R

(c)cosec

2

θ − cot

2

θ = 1 ;

cosec

θ ≥ 1 ∀ θ ∈ R

2. I

MPORTANT

T

′ R

ATIOS

:

(a)sin n

π = 0 ;

cos n

π = (-1)

n

_;

_{tan n}

π = 0

_{where n ∈ I}

(b)sin

2 )

1 n

2 (

+

π

= (−1)

n

_&cos

2 )

1 n

2 (

+

π

= 0 where n ∈ I

(c)sin 15° or sin

12 π

=

2

1 3−

= cos 75° or cos

12 5π

;

cos 15° or cos

12 π

=

2

1 3+

= sin 75° or sin

12 5π

;

tan

15° =

1

3

1

3 +

−

=

2− 3

= cot 75° ; tan

75° =

1

3

1

3 −

+

=

2+ 3

= cot 15°

(d)sin

8 π

=

2

2 2−

; cos

8 π

=

2

2 2+

; tan

8 π

=

2− ; tan

1 3π

₈

=

2+

1 (e) sin

10 π

or sin 18° =

4

1 5−

& cos 36° or cos

5 π

=

4

1 5+

3. T

RIGONOMETRIC

F

UNCTIONS

O

F

A

LLIED

A

NGLES

:

If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called A

LLIED

A

NGLES

.

(a)

sin (−

θ) = − sin

θ

; cos (−

θ) = cos

θ

(b)

_{sin (90°-}

θ) = cos

θ

_{; cos (90°}

−

θ) = sin

θ

(c) sin (90°+

θ) = cos

θ ; cos (90°+

θ) = − sin

θ (d)sin (180°−

θ) = sin

θ; cos (180°−

θ) = − cos

θ

(e)

_{sin (180°+}

θ) = − sin

θ ; cos (180°+

θ) = − cos

θ

(f) sin (270°−

θ) = − cos

θ ; cos (270°−

θ) = − sin

θ (g) sin (270°+

θ) = − cos

θ ; cos (270°+

θ) = sin

θ

4. T

RIGONOMETRIC

F

UNCTIONS

O

F

S

UM

O

R

D

IFFERENCE

O

F

T

WO

A

NGLES

:

(a) sin (A ± B) = sinA cosB ± cosA sinB

(b) cos (A ± B) = cosA cosB

∓

sinA sinB

(c)

sin²A − sin²B = cos²B − cos²A = sin (A+B) . sin (A− B)

(d)

cos²A − sin²B = cos²B − sin²A = cos (A+B) . cos (A

− B)

(e)

_tan

_{(A ± B) =}

tan tan

tan tan A B A B ± 1 ∓

(f) cot (A ± B) =

cot cot cot cot A B B A ∓1 ±

5. F

ACTORISATION

O

F

T

HE

S

UM

O

R

D

IFFERENCE

O

F

T

WO

SINES

O

R

COSINES

:

(a) sinC + sinD = 2 sin

2 D

C+

cos

2 D

C−

(b) sinC − sinD = 2 cos

2 D

C+

sin

2 D

C−

(c) cosC + cosD = 2 cos

2 D

C+

cos

2 D

C−

(d) cosC − cosD = −

2 sin

2 D

C+

sin

2 D

C−

6. T

RANSFORMATION

O

F

P

RODUCTS

I

NTO

S

UM

O

R

D

IFFERENCE

O

F

SINES

&

COSINES

:

(a) 2 sinA cosB = sin(A+B) + sin(A−B)

(b) 2 cosA sinB = sin(A+B) − sin(A−B)

(c) 2 cosA cosB = cos(A+B) + cos(A−B)

(d) 2 sinA sinB = cos(A−B) − cos(A+B)

7. M

ULTIPLE

A

NGLES

A

ND

H

ALF

A

NGLES

:

(a)

_{sin 2A = 2 sinA cosA ; sin}

θ = 2 sin

θ

2

cos

θ 2 13

(16)

(b)

cos2A = cos

2

_{A − sin}

2

_{A = 2cos}

2

_A

−

_{1 = 1 − 2 sin}

2

_{A ;}

cos

θ = cos

2θ 2

− sin²

θ 2

= 2cos

2θ 2

−

1 = 1 − 2sin

2θ 2

.

2 cos

2

_{A = 1 + cos 2A , 2sin}

2

_{A = 1 − cos 2A ; tan}

2

_{A =}

A

2 cos

1 A

2 cos

1 +

−

2 cos

2

2 θ

= 1 + cos

θ , 2 sin

2

2 θ

= 1 − cos

θ.

(c)

_{tan 2A =}

A

tan

1 A

tan

2

−

; tan

θ =

1 tan

(

2 )

)

2 (

tan

2

_θ

−

θ

(d)

sin 2A =

A

tan

1 A

tan

2

+

, cos 2A =

₁

_tan

_A

A

tan

1

2 2

+

−

(e)

sin 3A = 3 sinA

− 4 sin

3

_A

(f)

cos 3A = 4 cos

3

_{A − 3 cosA}

_(g)

_{tan 3A =}

A

tan

3

1 A

tan

A

tan

3

2 3

−

8. T

HREE

A

NGLES

:

(a)

tan

(A+B+C) =

A

tan

C

tan

C

tan

B

tan

B

tan

A

tan

1 C

tan

B

tan

A

tan

C

tan

B

tan

A

tan

−

+

N

OTE

I

F

:

(i)

A+B+C = π then tanA + tanB + tanC = tanA tanB tanC

(ii)

A+B+C =

2 π

then tanA tanB + tanB tanC + tanC tanA = 1

(b)

If A + B + C = π then :

(i)

sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii)

sinA + sinB + sinC = 4 cos

2 A

cos

2 B

cos

2 C

9. M

AXIMUM

& M

INIMUMVALUESOF

T

RIGONOMETRIC

F

UNCTIONS

:

(a)

Min. value of a

2

_tan

2

θ + b

2

_cot

2

θ = 2ab where θ ∈ R

(b)

Max. and Min. value of acosθ + bsinθ are

2 2

b

a +

and –

a +

2

b

2

(c)

If f(θ) = acos(α + θ) + bcos(β + θ) where a, b, α and β are known quantities then

–

_a

2

₊

_b

2

₊

₂

_ab

_cos(

_α

₋

_β

₎

_{< f(θ) <}

)

cos(

ab

2 b

a

2

+

2

+

α

−

β

(d)

If α,β ∈

0

2 ,

π













_{and α + β = σ (constant) then the maximum values of the expression}

cosα cosβ, cosα + cosβ, sinα + sinβ and sinα sinβ

occurs when α = β = σ/2.

(e)

If α,β ∈

0

2 ,

π









_



and α + β = σ(constant) then the minimum values of the expression

secα + secβ, tanα + tanβ, cosecα + cosecβ occurs when α = β = σ/2.

(f)

If A, B, C are the angles of a triangle then maximum value of

sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 60

0

(g)

In case a quadratic in sinθ or cosθ is given then the maximum or minimum values can be interpreted

by making a perfect square.

10. Sum of sines or cosines of

n angles,

sin

α + sin

(α + β) + sin

(α + 2β ) + ... + sin

(

α+ −n 1β

)

₌

2 2 n

sin

β β

sin



_



α

+

−

β



_



2

1 n

cos

α + cos

(α + β) + cos

(α + 2β ) + ... + cos

(

α+ −n 1β

)

=

2 2 n

sin

β β

cos













β

−

+

α

2

1 n

EXERCISE–I

Q.1 Prove that cos²α + cos² (α + β) − 2cos α cos β cos (α + β) = sin²β

Q.2 Prove that cos 2α = 2 sin²β + 4cos (α + β) sin α sin β + cos 2(α + β)

Q.3 Prove that , tan

α + 2 tan

2α + 4 tan

4α + 8 cot

8 α = cot

α .

Q.4 Prove that :

(a) tan 20° . tan 40° . tan 60° . tan 80° = 3

(b) tan

9° − tan

27° − tan

63° + tan

81° = 4 . (c)

2

3

16

7 sin

16

5 sin

16

3 sin

16 sin

4

π

+

4

π

+

4

π

+

4

π

=

Q.5 Calculate without using trigonometric tables :

(a) cosec 10°

−

3

sec 10°

(b) 4 cos 20° −

3

cot 20° (c)

°

−

_°

°

20 sin

20 cos

40 cos

2 (d)













°

−

°

+

°

2 sin

35

5 sin

40 cos

2

5 sec

10 sin

2

2 (e) cos

6

16 π

+ cos

6

16 3π

+ cos

6

16 5π

+ cos

6

16 7π

(f) tan 10° − tan 50° + tan 70°

Q.6(a) If X = sin













_θ

₊

π

12

7 + sin













_θ

₋

π

12 + sin













_θ

₊

π

12

3 , Y = cos













_θ

₊

π

12

7 + cos













_θ

₋

π

12 + cos













_θ

₊

π

12

3

14

(17)

then prove that

X

Y

X − = 2 tan2θ.

(b) Prove that sin²12° + sin² 21° + sin² 39° + sin² 48° = 1+ sin² 9° + sin² 18° .

Q.7 Show that :

(a)

cot 7

1

2 °

or tan 82

1 2 °

=

(

3 +

2 )( )

2 +

1 or

2 +

3 +

4 +

6 (b)

tan

142

1 2 °

= 2 +

2 − 3− 6

.

Q.8 If m tan (θ - 30°) = n tan (θ + 120°), show that cos 2

θ =

)

n

m

(

2 n

m

−

+

.

Q.9 If tan











 +

π

2 y

4 = tan

3

_









 +

π

2 x

4 , prove that

sin

x

y

sin

=

x

sin

3

1 x

sin

3

2 2

+

.

Q.10 If cos (α + β) =

4 5

; sin (α - β) =

5

13

& α , β lie between 0 &

₄

π

, then find the value of tan

2α.

Q.11 Prove that if the angles α & β satisfy the relation

₍

₎

(

m

n

)

m

n

2 sin

sin

₌

_>

β

+

α

β

then

n

m

tan

1 n

m

1

_tantan

−

β

α

−

+

=

α β

.

Q.12 (a) If y = 10 cos²x − 6 sin x cos x + 2 sin²x , then find the greatest & least value

of y .

(b) If y = 1

+

2 sin

x + 3

cos

2

_{x , find the maximum & minimum values of y ∀ x}

∈

_{R .}

(c) If y = 9 sec

2

_{x + 16 cosec}

2

_{x, find the minimum value of y ∀ x ∈ R.}

(d) Prove that 3 cos

_θ+ π  

3

+ 5 cos θ + 3 lies from - 4 & 10 .

(e) Prove that

(

2 3+4

)

sin θ + 4 cos θ lies between −

2 +

(

2

5 )

&

2 +

(

2

5 )

.

Q.13 If A + B + C = π, prove that

∑

_











C

tan

.

B

tan

A

tan

=

∑

(tan A)

A) − 2

∑

(cot A).

Q.14 If α + β = c where α, β > 0 each lying between 0 and π/2 and c is a constant, find the maximum or

minimum value of

(a)

sin α + sin β

(b)

sin α sin β

(c)

tan α + tan β

(d)

cosec α + cosec β

Q.15 Let A

₁

, A

₂

, ... , A

_n

be the vertices of an n-sided regular polygon such that ;

4 1 3 1 2 1

A

1 A

A

1 A

A

1 ₌

₊

. Find the value of n.

Q.16 Prove that : cosec

θ + cosec

2 θ + cosec

2

θ + ... + cosec

₂

n − 1

θ = cot

_{(θ/2) − cot 2}

n - 1

θ

Q.17 For all values of α , β , γ prove that;

cos α + cos β + cos γ + cos (α + β + γ) = 4 cos

2 β

+

α

.cos

2 γ

+

β

. cos

2 α

+

γ

.

Q.18 Show that

B

cos

A

cos

)

B

A

sin(

B

sin

2 A

sin

2 B

sin

1 B

cos

A

cos

A

sin

1 −

+

−

=

−

+

.

Q.19 If tan β =

tan tan

tan . tan

α γ

+ +

1

, prove that sin 2β =

sin sin sin . sin 2 2 1 2 2 α γ α γ + +

.

Q.20 If α + β = γ , prove that cos² α + cos² β + cos² γ = 1 + 2 cos α cos β cos γ .

Q.21 If α + β + γ =

π 2

, show that

(

)

(

)(

)

(

₂2

)

(

₂2

)(

2₂

)

tan

1 tan

1

γ β α γ β α

+

−

=

sin sin sin

cos cos cos

α β γ α β γ + + − + + 1

.

Q.22 If A + B + C

=

π and cot θ = cot A + cot B + cot C, show that ,

sin (A − θ) . sin (B − θ) . sin (C − θ) = sin

3

_{θ .}

Q.23 If P =

19

17 cos

...

19

5 cos

19

3 cos

19 cos

π

+

π

+

π

+

π

and

Q =

21

20 cos

...

21

6 cos

21

4 cos

21

2 cos

π

+

π

+

π

+

π

, then find P – Q.

Q.24 If A, B, C denote the angles of a triangle ABC then prove that the triangle is right angled if and only if

sin4A + sin4B + sin4C = 0.

Q.25 Given that (1 + tan 1°)(1 + tan 2°)...(1 + tan 45°) = 2

n

_{, find}

_n

_.

EXERCISE–II

Q.1 If tan

α = p/q where α = 6β, α being an acute angle, prove that;

2

1 (p

cosec 2

β − q

sec 2

β) =

2 2

q

p +

.

Q.2 Let A

₁

, A

₂

, A

₃

... A

_n

are the vertices of a regular n sided polygon inscribed in a circle of radius R.

If (A

₁

A

₂

)

2

_{+ (A}

(18)

Q.3 Prove that:

1 )

cos(

2

3 cos

−

φ

−

θ

φ

+

θ

= (cosθ + cosφ) cos(θ + φ) – (sinθ + sinφ) sin(θ + φ)

Q.4 Without using the surd value for sin 18

0

_{or cos 36}

0

_{, prove that 4 sin 36}

0

_{cos 18}

0

_{= 5}

Q.5 Show that ,

2

1 x

27 cos

x

9 sin

x

9 cos

x

3 sin

x

3 cos

x

sin

₊

₌

(tan27x − tanx)

Q.6 Let x

₁

=

∏

=

π

5 1 r

11 r

cos

and x

₂

=

∑

=

π

5 1 r

11 r

cos

, then show that

x

₁

· x

₂

=

64

1 _











−

π

1

22 ec

cos

, where Π

Π

denotes the continued product.

Q.7 If θ =

7 2π

, prove that tan

θ . tan 2

θ + tan 2

θ . tan 4

θ + tan 4

θ . tan

θ = −

7.

Q.8 For 0 < x <

π

4

prove that ,

sin

x

(cos

x

sin

x

)

x

cos

2

₋

> 8.

Q.9 (a) If α =

7 2π

prove that, sin

α

+ sin

2α + sin

4α =

2

7 (b) sin

7 π

. sin

7 2π

. sin

7 3π

=

8

7 Q.10 Let k = 1°, then prove that

∑

=

+

88 0 n

cos

nk

· cos(

n

1 )

k

1 =

k

sin

k

cos

2

Q.11 Prove that the value of cos

A

+ cos

B

+ cos

C lies between 1 &

2

3 where A

A

+

B

+

C = π.

Q.12 If cosA = tanB, cosB = tanC and cosC = tanA , then prove that sinA = sinB = sinC = 2 sin18°.

Q.13 Show that

x

sin

x

cos

3 +

R

x∈

∀

can not have any value between

−

2

2 and

2

2 . What inference

can you draw about the values of

x

cos

3 x

sin

+

?

Q.14 If (1 + sin t)(1 + cos t) =

4

5 . Find the value of (1 – sin t)(1 – cos t).

Q.15 Prove that from the equality

b

a

1 b

cos

a

sin

4 4

+

=

α

+

α

follows the relation ;

3

_{( )}

3 8 3 8

b

a

1 b

cos

a

sin

+

=

α

+

α

.

Q.16 Prove that the triangle ABC is equilateral iff , cot

A + cot

B + cot

C =

3

.

Q.17 Prove that the average of the numbers

n

sin

n

°,

n

= 2, 4, 6, ..., 180, is cot 1°.

Q.18 Prove that : 4 sin

27° =

( ) ( )

₅

+

₅

1/2

−

₃

−

₅

1/2

.

Q.19 If A+B+C = π; prove that tan

2

2 A

+ tan

2

2 B

+ tan

2

2 C

_{≥ 1.}

Q.20 If A+B+C = π (A , B , C > 0) , prove that sin

2 A

. sin

2 B

. sin

2 C

_≤

8

1 .

Q.21 Show that elliminating x & y from the equations , sin

x + sin

y = a ;

cos

x + cos

y = b & tan

x + tan

y = c gives

(

₂ ₂

)

2 ₂

a

4 b

a

b

a

8 −

+

= c.

Q.22 Determine the smallest positive value of x (in degrees) for which

tan(x + 100°) = tan(x + 50°) tan x tan (x – 50°).

Q.23 Evaluate :

∑

= − − n 1 n 1 n 1 n n

2 x

cos

2

2 x

tan

Q.24 If

α

+

β

+

γ

=

π

&













α

+

β

−

γ













γ

+

α

−

β













β

+

γ

−

α

4 tan

·

4 tan

·

4 tan

= 1, then prove that;

1 + cos

α

+ cos

β

+ cos

γ

_{= 0.}

Q.25 ∀

x

∈

R, find the range of the function, f (x) = cos x (sin x + sin

2

x +

sin

2

α ) ;

α

∈

[0,

π

]

EXERCISE–III

Q.1 sec

2

θ

₌

2

)

y

x

(

xy

4 +

is true if and only if :

[JEE ’96, 1]

(A) x + y

≠

0 (B) x = y , x

≠

0 (C) x = y

(D) x

≠

0 , y

≠

0

Q.2 (a)

Let n be an odd integer. If sin n

θ

=

r n =

∑

0

b

r

sin

r

θ

_{, for every value of}

θ

_{, then :}

(19)

(C) b

₀

=

−

1, b

₁

= n

(D) b

₀

= 0, b

₁

= n

2

−

_{3n + 3}

(b)

Let A

₀

A

₁

A

₂

A

₃

A

₄

A

₅

be a regular hexagon inscribed in a circle of unit radius .

Then the product of the lengths of the line segments

A

₀

A

₁

, A

₀

A

₂

& A

₀

A

₄

is :

(A)

3

4

(B) 3

3

(C) 3

(D)

3 3 2

(c)

Which of the following number(s) is/are rational ? [ JEE '98, 2

+

2 +

2 = 6 out of 200 ]

(A) sin 15º

(B) cos 15º

(C) sin 15º cos 15º

(D) sin 15º cos 75º

Q.3 For a positive integer n, let f

_n

(

θ

) =













θ

2 tan

(1+ sec

θ

)

(1+ sec

2 θ

)

(1+

sec

4 θ

)

....

(1

+ sec2

n

θ

_{) Then}

(A) f

₂_₁₆π_

= 1

(B) f

₃ _₃₂π_

= 1

(C) f

₄_₆₄π_

= 1 (D) f

₅_₁₂₈π _

= 1 [JEE '99,3]

Q.4(a) Let f

(

θ

) = sin

θ

(sin

θ

+ sin 3

θ

) . Then f

(

θ

)

: [ JEE 2000 Screening. 1 out of 35 ]

(A)

≥

0 only when

θ

≥

0 (B)

≤

0 for all real

θ

(C)

≥

0 for all real

θ

(D)

≤

0 only when

θ

≤

0 .

(b) In any triangle ABC, prove that, cot

2 A

+ cot

2 B

+ cot

2 C

= cot

2 A

cot

2 B

cot

2 C

. [JEE 2000]

Q.5(a) Find the maximum and minimum values of 27

cos 2x

_{· 81}

sin 2x

_.

(b) Find the smallest positive values of

x & y satisfying, x

−

y =

4 π

, cot x + cot y = 2. [REE 2000, 3]

Q.6 If

α

+

β

=

π

2 and

β + γ = α then tanα equals

[ JEE 2001 (Screening), 1 out of 35 ]

(A) 2(tanβ + tanγ)

(B) tanβ + tanγ

(C) tanβ + 2tanγ

(D) 2tanβ + tanγ

Q.7 If θ and φ are acute angles satisfying sinθ =

2

1 , cos φ =

3

1 , then θ + φ ∈ [JEE 2004 (Screening)]

(A)



_



π

_



2 ,

3 (B)













π

3

2 ,

2 (C)













π

6

5 ,

3

2 (D)



_



5 ₆

π

,

π



_



Q.8 In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they

touch each other and also the sides of the triangle. Area of the triangle is

(A) 4 + 2 3

(B) 6 + 4 3

(C) 12 +

4

3

7 (D) 3 +

4

3

7 [JEE 2005 (Screening)]

Q.9 Let θ ∈



_



π



_



4 ,

0 _{and t}

1

= (tanθ)

tanθ

, t

2

= (tanθ)

cotθ

, t

3

= (cotθ)

tanθ

, t

4

= (cotθ)

cotθ

, then

(A) t

₁

> t

₂

> t

₃

> t

₄

(B) t

₄

> t

₃

> t

₁

> t

₂

(C) t

₃

> t

₁

> t

₂

> t

₄

(D) t

₂

> t

₃

> t

₁

> t

₄

[JEE 2006, 3]

ANSWER SHEET

(

EXERCISE–I)

Q 5. (a) 4

(b) −1 (c)

3

(d) 4 (e)

4

5 (f)

3

Q 10.

56 33

Q 12. (a) y

_max

= 11 ; y

_min

= 1 (b) y

_max

=

13

3 ; y

min

= −

1, (c) 49

Q14.

_{(a) max = 2 sin (c/2), (b) max. = sin}

2

_{(c/2), (c) min = 2 tan (c/2), (d) min = 2 cosec (c/2)}

Q 15. n = 7

Q23.

1 Q.25 n = 23

EXERCISE –II

Q.2 _{n = 7 Q.13}











−

2

1 ,

2

1

Q.14

10

4 13 −

Q.22 _{x = 30°}

Q 23.

1 n 1 n

2 x

sin

2

1 x

2 sin

2

− −

−

Q.25 _{– 1}

_{+ sin α ≤ y ≤ 1}

2

_{+ sin α}

2

EXERCISE–III

Q.1 B

Q.2 (a) B, (b) C, (c) C

Q.3 A, B, C, D

Q.4 (a) C

Q.5 (a) max. = 3

5

_{& min. = 3}

–5

_{; (b) x =}

12 5π

; y =

6 π

Q.6 C

Q.7 B

Q.8 B

Q.9 B

17

(20)

EXERCISE–IV (Objective)

Part : (A) Only one correct option

1. 1. 1. 1.

( ) (

)

(

)

(

x

) (

.tan x

)

cos x sin x cos . x tan 2 3 2 2 7 2 3 2 3 + − − − + − π π π π π

when simplified reduces to:

(A) sinx cosx (B) −sin2 _x (C) −_sin_{x cos}_x _{(D) sin}2_x

2. The expression 3       α + π +       π₋_α ) 3 ( sin 2 3 sin4 4 _{– 2}       α + π +      π₊_α ) 5 ( sin 2 sin6 6 _{is equal to}

(A) 0 (B) 1 (C) 3 (D) sin 4α + sin 6α

3. If tanA & tanB are the roots of the quadratic equation x2 − ax + b = 0, then the value of sin2 _{(A + B).}

(A) 2 2 2 ) b 1 ( a a − + (B) 2 2 2 b a a + (C) 2 2 ) c b ( a + (D) 2 2 2 ) a 1 ( b a − 4. The value of log₂ [cos2 (α + β) + cos2 (α − β) − cos 2α. cos 2β] :

(A) depends on α & β both (B) depends on α but not on β

(C) depends on β but not on α (D) independent of both α & β.

5. _° ° ° ° + ° 80 sin 10 sin 50 sin 70 sin 8 20 cos 2 is equal to: (A) 1 (B) 2 (C) 3/4 (D) none

6. If cos A = 3/4, then the value of 16cos2_{(A/2) – 32 sin (A/2) sin (5A/2) is}

(A) – 4 (B) – 3 (C) 3 (D) 4

7. If y = cos2 (45º + x) + (sin x − cos x)2_{then the maximum & minimum values of y are:}

(A) 2 & 0 (B) 3 & 0 (C) 3 & 1 (D) none

8. The value of cos

19 π + cos 19 3π + cos 19 5π +... + cos 19 17π is equal to: (A) 1/2 (B) 0 (C) 1 (D) none

9. The greatest and least value of log ₂

(

sinx−cosx+3 2

)

are respectively:

(A) 2 & 1 (B) 5 & 3 (C) 7 & 5 (D) 9 & 7

10. In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the opposite

vertex. Then the other acute angles of the triangle are (A) 3 π & 6 π (B) 8 π & 8 3π (C) 4 π & 4 π (D) 5 π & 10 3π 11. _cos₂₉₀1 _° + ₃_sin₂₅₀_° 1 = (A) 3 3 2 (B) 3 3 4 (C) 3 (D) none 12. If 4 3π < α < π, then α + α ₂ sin 1 cot 2 is equal to

(A) 1 + cot α (B) – 1 – cot α (C) 1 – cot α (D) – 1 + cot α

13. If x ∈      _π π 2 3 , then 4 cos2       − π 2 x 4 + 4sin x sin 2x 2 4 ₊ _{is always equal to}

(A) 1 (B) 2 (C) – 2 (D) none of these

14. If 2 cos x + sin x = 1, then value of 7 cos x + 6 sin x is equal to

(A) 2 or 6 (B) 1 or 3 (C) 2 or 3 (D) none of these

15. If cosec A + cot A = 2 11 , then tan A is (A) 22 21 (B) 16 15 (C) 117 44 (D) 43 117

16. If cot α + tan α = m and

α cos 1 – cos α = n, then (A) m (mn2₎1/3_{– n(nm}2₎1/3_{= 1} _{(B) m(m}2_n)1/3_{– n(nm}2₎1/3_{= 1} (C) n(mn2₎1/3_{– m(nm}2₎1/3_{= 1} _{(D) n(m}2_n)1/3_m(mn2₎1/3_{= 1}

17. The expression cos_cos6x₅+_x6₊cos₅_cos4x₃+_x15₊cos₁₀_cos2x_x+10 is equal to

(A) cos 2x (B) 2 cos x (C) cos2_x _{(D) 1 + cos x}

18. If B sin A sin = 2 3 and B cos A cos = 2 5

, 0 < A, B < π/2, then tan A + tan B is equal to

(A) 3/ 5 (B) 5/ 3 (C) 1 (D) ( 5+ 3)/ 5

19. If sin 2θ = k, then the value of ₊ θ2_θ

3 tan 1 tan + ₊ θ2_θ 3 cot 1 cot is equal to (A) 1 k 2 − (B) 2 k 2 − (C) k182_{+ 1} _{(D) 2 – k}2

(21)

Part : (B) May have more than one options correct 20. Which of the following is correct ?

(A) sin 1° > sin 1 (B) sin 1° < sin 1 (C) cos 1° > cos 1 (D) cos 1° < cos 1 21. If 3 sin β = sin (2α + β), then tan (α + β) – 2 tan α is

(A) independent of α (B) independent of β

(C) dependent of both α and β (D) independent of α but dependent of β

22. It is known that sinβ =

5 4

& 0 < β < π then the value of

α β + α − β + α _π sin ) ( cos ) ( sin 3 6 cos 2 is:

(A) independent of α for all β in (0, π) (B)

3 5 for tan β > 0 (C) 15 ) cot 24 7 ( 3 + α

for tanβ < 0 (D) none

23. If the sides of a right angled triangle are {cos2α + cos2β + 2cos(α + β)} and {sin2α + sin2β + 2sin(α + β)}, then the length of the hypotenuse is:

(A) 2[1+cos(α − β)] (B) 2[1 − cos(α + β)] (C) 4 cos2

2 β − α (D) 4sin2 2 β + α 24. If x = secφ − tanφ & y = cosecφ + cotφ then:

(A) x = y_y+₋₁1 (B) y = 1₁₋+ x_x (C) x = y_y−₊₁1 (D) xy + x − y + 1 = 0

25. (a + 2) sin α + (2a – 1) cos α = (2a + 1) if tan α = (A) 4 3 (B) 3 4 (C) 1 a a 2 2₊ (D) 1 a a 2 2₋ 26. If tan x = c a b 2 − , (a ≠ c)

y = a cos2_{x + 2b sin x cos x + c sin}2_x

z = a sin2_{x – 2b sin x cos x + c cos}2_{x, then}

(A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2_{+ 4b}2 27. n B sin A sin B cos A cos _      − + + n B cos A cos B sin A sin _      − + (A) 2 tann 2 B A− (B) 2 cotn 2 B A−

: n is even (C) 0 : n is odd (D) none

28. The equation sin6_{x + cos}6_{x = a}2_{has real solution if}

(A) a ∈ (–1, 1) (B) a ∈       − − 2 1 , 1 _{(C) a ∈}       − 2 1 2 1 (D) a ∈       1 , 2 1

EXERCISE–IV (Subjective)

1. The minute hand of a watch is 1.5 cm long. How far does its tip move in 50 minutes?

(Use π = 3.14).

2. If the arcs of the same length in two circles subtend angles 75° and 120° at the centre, find the ratio of their radii.

3. Sketch the following graphs :

(i) y = 3 sin 2x (ii) y = 2 tan x (iii) y = sin

2 x

4. Prove that cos _3₂π+θ_ cos (2π + θ) 

     θ + π +       π₋_θ ) 2 ( cot 2 3 cot _{= 1.}

5. Prove that cos 2 θ cos

2 θ – cos 3 θ cos 2 9θ = sin 5 θ sin 2 5θ . 6. If tan x = 4 3 , π < x < 2 3π

, find the value of sin 2 x and cos 2 x . 7. prove that               α α +      α−π +      α−π − 4 cot 2 cos 4 cot 1 4 cot 1 2 2 sec 2 9α = cosec 4α. 8. Prove that, sin 3x. sin3_{x + cos 3}_{x. cos}3_{x = cos}3₂_x.

9. If tanα =

q p

where α = 6β, α being an acute angle, prove that; 2 1

(pcosec 2β − qsec 2β) = _p2₊_q2 _.

10. If tan β =₁tan₊_tanα+_α_.tan_tanγ_γ , prove that sin 2β =₁sin₊_sin2α₂+_α_.sin_sin2₂γ_γ.

11. Show that: (i) cot 7

2 1° or tan 82 2 1° =

(

3+ 2

)( )

2+1 or 2+ 3+ 4+ 6 (ii) tan142 2 1° = 2 + 2− 3− 6. 19 (iii) 4 sin27° =

(

5+ 5

) (

1/2− 3− 5

)

1/2

(22)

12. Prove that, tanα + 2 tan2α + 4 tan4α + 8 cot8 α = cotα. 13. If cos (β − γ) + cos (γ − α) + cos (α − β) =

2 3 −

, prove that cos α + cos β + cos γ = 0, sin α + sin β + sin γ = 0.

14. Prove that from the equality

b a 1 b cos a sin4 4 + = α + α

follows the relation ₃

₍

₎

₃

8 3 8 b a 1 b cos a sin + = α + α

15. Prove that: cosecθ + cosec2θ + cosec22 θ +... + cosec₂ n − 1θ = cot(θ/2) − cot 2n − 1 θ. Hence or

otherwise prove that cosec 15 4π +cosec 15 8π +cosec 15 16π + cosec 15 32π = 0 16. Let A1, A2,..., An be the vertices of an n−sided regular polygon such that;

4 1 3 1 2 1 A A 1 A A 1 A A 1 ₌ ₊ . Find the value of n.

17. If A + B + C = π, then prove that

(i) tan² 2 A + tan² 2 B + tan² 2 C ≥ 1 (ii) sin 2 A . sin 2 B . sin 2 C ≤ 8 1 .

(iii) cos A + cos B + cos C ≤

2 3 18. If _θ cos ax + _θ sin by = a2_{– b}2_, θ θ 2 cos sin ax – θ θ 2 sin cos by

= 0. Show that (ax)2/3_{+ (by)}2/3_{= (a}2_{– b}2₎2/3

19. If P_n = cosnθ + sinnθ and Q n = cos

nθ – sinnθ, then show that

P_n – P_{n – 2} = – sin2θ cos2θ P n – 4

Q_n – Q_{n – 2}= – sin2θ cos2θ Q

n – 4 and hence show that

P₄ = 1 – 2 sin2θ cos2θ , Q 4 = cos

2θ – sin2θ

20. If sin(θ + α) = a & sin(θ + β) = b (0 < α, β, θ < π/2) then find the value of cos2(α − β) − 4ab cos(α − β)

21. If A + B + C = π, prove that

tan B tan C + tan C tan A + tan A tan B = 1 + sec A. sec B. sec C.

22. If tan2α + 2tanα. tan2β = tan2β + 2tanβ. tan2α, then prove that each side is equal to 1 or

tan α = ± tan β.

EXERCISE–IV

1. D 2. B 3. A 4. D 5. B 6. C 7. B 8. A 9. B 10.B 11. B 12. B 13. B 14.A 15. C 16.A 17.B 18.D 19. B 20. BC 21. AB 22. BC 23. AC 24. BCD 25.BD 26. BC 27. BC 28. BD

EXERCISE–V

1. 7.85 cm 2. r₁ : r₂ = 8 : 5 6. sin 2 x = 10 3 and cos 2 x = – 10 1 16.n = 7 20.1